Since time=distance/speed, if his speed is s and the current's is c, then
15/(s+c) = 15/(s-c)-5
15/(2s+c) = 15/(2s-c)-1
Now just solve for s and c.
At an ordinary rate, a man can row the distance from A to B, about 15 km, in 5 hours less time than it takes him to return. Could he double his rate, his time to B would only be one hour less than his time to A. What is his usual rate in still water? What is the rate of the stream?
6 answers
My working equations were 15/(s+c) = 5 - 15/(s-c) and 15/(2s+c) = 15/(2s-c) - 1
Whichever I use, I can't seem to get the right values of s and c. My answers are not in the choices given.
Whichever I use, I can't seem to get the right values of s and c. My answers are not in the choices given.
I don't think Steve is on line right now, so I will try.
Did you notice the the right side of both of Steve's equations are the opposite of yours?
Steve's solution would result in
s = 4
c = 2
Yours give inadmissible results of
s = appr -14 and c = 16.7
Your translation of the given data into equations is incorrect.
Did you notice the the right side of both of Steve's equations are the opposite of yours?
Steve's solution would result in
s = 4
c = 2
Yours give inadmissible results of
s = appr -14 and c = 16.7
Your translation of the given data into equations is incorrect.
Oh yeah, I see that.
Well, I guess I'm a bit confused with the less thingy. Anyway, could you please explain how you got 4&2?
Thank you.
Well, I guess I'm a bit confused with the less thingy. Anyway, could you please explain how you got 4&2?
Thank you.
Nevermind, I got it. Thanks for all the help!
How were you able to obtain the answer? My equation is similar to what steve formulated but im stuck in the problem thanks for the help.